Question

Suppose the gravitational force varies inversely as the ${n^{th}}$ power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be
A. ${R^{\left( {n + 1} \right)/2}}$
B. ${R^{\left( {n - 1} \right)/2}}$
C. ${R^n}$
D. ${R^{(n - 2)/2}}$

Answer 1

To solve this question we should know what is called gravitation and gravitational force. Not only for the earth but also for all the planet's gravity comes from its masses. It is a force which is exerted by the earth on all the objects out there. When any object is thrown up it falls to the ground. This happens due to the gravitational force. we have to remember these things.

Complete Step by Step Answer:
Here we know that, F $= K{R^{ - n}}$
$F$ $= MR{\omega ^2}$
We know that, ${\omega ^2} = K{R^{ - n + 1}}$
$\omega = K\dfrac{{{R^{ - \left( {n + 1} \right)}}}}{2}$
$\Rightarrow\dfrac{{2\pi }}{T} \propto {R^{ - \dfrac{{n + 1}}{2}}}$
$\therefore T \propto {R^{ + \dfrac{{n + 1}}{2}}}$

Therefore the right option would be A.

Note: We know that, here R is equal to radius and F is equal to force. According to the question, we know that the mass is equal to M. We have calculated the time period with the help of mass, force which is centripetal force. We can get confused by the force and the centripetal force. For that we should know the definition of centripetal force. a force that is responsible for moving the body which is moving in a circular path and is directed towards the centre around which the body is moving. This force is responsible for the motion of the earth which is moving around the sun without any restriction. This is an external force created by the sun. Another simple example of this force is roller skates and a ring floor.